# Derivatives of Inverse Trigonometric Functions

In this section, we will learn the derivative formulas of the inverse trigonometric functions with their proofs.

Now we calculate the derivative of $\sin^{-1}x$ (or arc sin x).

#### Derivative of $\sin^{-1}x$

$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} \, (|x|<1)$

Proof: Note that $\sin^{-1}x$ is defined when $|x| \leq 1.$ We assume that
$y=\sin^{-1}x.$ As $|x| \leq 1$, we must have that $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.$ $y=\sin^{-1}x \Rightarrow x=\sin y$ Differentiating with respect to $y$, we have $\frac{dx}{dy}=\frac{d}{dx}(\sin y)=\cos y$ $\Rightarrow \frac{dx}{dy}=1-\sin^2 y$ $\Rightarrow \frac{dx}{dy}=\sqrt{1-x^2}$ Therefore, we deduce that $\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{\sqrt{1-x^2}}$ So the derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$, where $|x|\leq 1.$ ♣

Next, we find the derivative of $\cos^{-1}x$ (or arc cos x).

#### Derivative of $\cos^{-1}x$

$\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}} \, (|x|<1)$

Proof: We know that $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ Differentiating with respect to $x$, we get $\frac{d}{dx}(\sin^{-1}x+\cos^{-1}x)=\frac{d}{dx}(\frac{\pi}{2})$$\Rightarrow \frac{d}{dx}(\sin^{-1}x)+\frac{d}{dx}(\cos^{-1}x)=0$

[as the derivative of a constant is zero]

$\Rightarrow \frac{1}{\sqrt{1-x^2}}+\frac{d}{dx}(\cos^{-1}x)=0$ $\Rightarrow \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}$ So the derivative of $\cos^{-1}x$ is $-\frac{1}{\sqrt{1-x^2}}$, where $|x|\leq 1.$ ♣

Let us now calculate the derivative of $\tan^{-1}x$ (or arc tan x).

#### Derivative of $\tan^{-1}x$

$\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} \, (-\infty<x<\infty)$

Proof: Note that $\tan^{-1}x$ is defined when $-\infty<x<\infty.$ We assume that $y=tan^{-1}x.$ As $-\infty<x<\infty$, we must have that $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.$ $y=\tan^{-1}x \Rightarrow x=\tan y$ Differentiating with respect to $y$, we have $\frac{dx}{dy}=\frac{d}{dx}(\tan y)=\sec^2 y$ $\Rightarrow \frac{dx}{dy}=1+\tan^2 y$ $\Rightarrow \frac{dx}{dy}=1+x^2$ Therefore, we deduce that $\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{1+x^2}$ So the derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, where $-\infty<x<\infty.$ ♣

Next, we find the derivative of $\cot^{-1}x$ (or arc cot x).

#### Derivative of $\cot^{-1}x$

$\frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1+x^2} \, (-\infty<x<\infty)$
Proof:  We know that $\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$ Differentiating with respect to $x$, we get $\frac{d}{dx}(\tan^{-1}x+\cot^{-1}x)=\frac{d}{dx}(\frac{\pi}{2})$ $\Rightarrow \frac{d}{dx}(\tan^{-1}x)+\frac{d}{dx}(\cot^{-1}x)=0$

[as the derivative of a constant is zero]

$\Rightarrow \frac{1}{1+x^2}+\frac{d}{dx}(\cot^{-1}x)=0$ $\Rightarrow \frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1-x^2}$ So the derivative of $\cot^{-1}x$ is $-\frac{1}{1+x^2}$, where $-\infty<x<\infty.$ ♣

We now calculate the derivative of $\sec^{-1}x$ (or arc sec x).

#### Derivative of $\sec^{-1}x$

$\frac{d}{dx}(\sec^{-1}x)=\frac{1}{x\sqrt{x^2-1}} \, (|x|>1)$
Proof: Note that $\sec^{-1}x$ is defined when $|x|>1.$ We assume that $y=\sec^{-1}x.$ $\Rightarrow x=\sec y$ Differentiating with respect to $y$, we have $\frac{dx}{dy}=\frac{d}{dx}(\sec y)=\sec y \tan y$ $\Rightarrow \frac{dx}{dy}=\sec y \sqrt{\sec^2y-1}$ $\Rightarrow \frac{dx}{dy}=x\sqrt{x^2-1}$ Therefore, we deduce that $\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{x\sqrt{x^2-1}}$ So the derivative of $\sec^{-1}x$ is $\frac{1}{x\sqrt{x^2-1}}$, where $|x|>1.$ ♣

Next, we find the derivative of $\text{cosec}^{-1}x$ (or arc cosec x).

#### Derivative of $\text{cosec}^{-1}x$

$\frac{d}{dx}(\text{cosec}^{-1}x)=-\frac{1}{x\sqrt{x^2-1}} \, (|x|>1)$
Proof: We know that $\sec^{-1}x+\text{cosec}^{-1}x=\frac{\pi}{2}$ Differentiating with respect to $x$, we get $\frac{d}{dx}(\sec^{-1}x+\text{cosec}^{-1}x)=\frac{d}{dx}(\frac{\pi}{2})$ $\Rightarrow \frac{d}{dx}(\sec^{-1}x)+\frac{d}{dx}(\text{cosec}^{-1}x)=0$

[as the derivative of a constant is zero]

$\Rightarrow \frac{1}{x\sqrt{x^2-1}}+\frac{d}{dx}(\text{cosec}^{-1}x)=0$ $\Rightarrow \frac{d}{dx}(\text{cosec}^{-1}x)=-\frac{1}{x\sqrt{x^2-1}}$ So the derivative of $\text{cosec}^{-1}x$ is $-\frac{1}{x\sqrt{x^2-1}}$, where $|x|> 1.$ ♣

Share via: